Question

The value of $1^2-2^2+3^2-4^2+5^2-6^2+\dots n\text{terms}$ is/are

• A. $\frac{n}{2},$ when $n$ is odd
• B. $-\frac{(n+1)}{2},$ when $n$ is even
• C. $-\frac{n(n+1)}{2},$ when $n$ is even
• D. $\frac{n(n+1)}{2},$ when $n$ is odd

Answer 1

Answer: (C), (D)

$\frac{n(n+1)}{2},$ when $n$ is odd

Clearly, $n^{th}$ term of the given series is negative or positive accordingly as $n$ is even or odd, respectively.

Case $1:n$ is even
$1^2-2^2+3^2-4^2+5^2-6^2+\dots +(n-1{)}^2-n^2$
$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+\dots +\left(\right(n-1{)}^2-n^2)$$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+\dots +\left(\right(n-1)-n)(n-1+n)$
$=-(1+2+3+4+\dots +(n-1)+n)$
$=-\frac{n(n+1)}{2}$

Case $2:n$ is odd
$(1^2-2^2)+(3^2-4^2)+\dots +\left\{\right(n-2{)}^2-(n-1{)}^2\}+n^2$
$=(1-2)(1+2)+(3-4)(3+4)+\dots +\left[\right(n-2)-(n-1\left)\right]\left[\right(n-2)+(n-1\left)\right]+n^2$
$=-(1+2+3+4+\dots +(n-2)+(n-1\left)\right)+n^2$
$=\frac{n(n+1)}{2}$